Montgomery, Alabama, Ranks No. 1 as Most Affordable for Home Buyers

Comparing 300 U.S. cities across 10 metrics, WalletHub names Montgomery, Alabama, No. 1 for affordability, followed by Flint, Michigan, and Toledo, Ohio.

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Montgomery, Alabama.

Adobe Stock / SeanPavonePhoto

Montgomery, Alabama.

As affordability becomes an even tighter squeeze, WalletHub has compared 300 U.S. cities across 10 metrics to find the most affordable places for home buyers. The study ranked Montgomery, Alabama, as the most affordable city for buyers.

The 10 metrics used to evaluate each city were housing affordability (median house price divided by median annual household income), maintenance affordability, average cost of homeowner’s insurance, cost of living, cost per square foot, real estate tax rate, rent-to-price ratio, median home price appreciation, quarterly active listings per capita, and vacancy rate.

The top 10 for overall affordability include:

  1. Montgomery, Alabama
  2. Flint, Michigan
  3. Toledo, Ohio
  4. Detroit
  5. Akron, Ohio
  6. Warren, Michigan
  7. Pittsburgh
  8. Yuma, Arizona
  9. Springfield, Illinois
  10. Palm Bay, Florida

In terms of the housing affordability metric alone, Springfield, Illinois, ranked No. 1, followed by Peoria, Illinois; Montgomery; Detroit; and Flint. The least affordable cities, according to WalletHub, include California’s Santa Barbara, Santa Monica, Berkeley, Glendale, and Los Angeles.

Looking at vacancy rates, Miami Beach, Florida, has the highest rate, followed by Tuscaloosa, Alabama; Detroit; Pompano Beach, Florida; and Flint. The lowest vacancy rates are in South Gate, California; Elk Grove, California; Fontana, California; and Arvada, Colorado.

About the Author

Leah Draffen

Leah Draffen is an associate editor at Builder. She earned a B.A. in journalism and minors in business administration and sociology from Louisiana State University.

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